Solving non-linear recurrence relation $p_{n+1} = \frac{2p_n}{p_n + 1}, p_1 \neq 0$ Given the following recurrence relation $p_{n+1} = \frac{2p_n}{p_n + 1}, p_1 \neq 0$, I want to show that as $n \rightarrow \infty, p_n \rightarrow 1$.
By assuming that such a limit exists, and setting $p_{n+1} \approx p_n$ for large $n$, I found that $p_n$ tends to 1.
Is there a more rigorous way of doing this?
Note: I just want to show that $p_n$ tends to 1, I don't need an explicit formula for $p_n$.
 A: Hint
Let $q_n=\frac 1 {p_n}$ to get something much more pleasant to work
A: Short answer.
Consider the sequence $q_n = \frac{1}{p_n} - 1$. Then $ q_{n+1} = \frac{1}{2}q_n $. So, we have $q_n \to 0$ as $n\to\infty$ provided $q_n \neq \infty$. This implies that $p_n \to 1$ as $n\to\infty$.
Long answer.
Assume that $f_A(z) = \frac{az+b}{cz+d}$ is a Möbius transform and $(p_n)$ is given by the iteration $p_{n+1} = f(p_n)$. (The issue of division by zero can be easily ignored by considering everything on the Riemann sphere $\mathbb{C}\cup\{\infty\}$.) We know that $f_A$ is associated to the matrix
$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. $$
Assume that $A$ is diagonalizable, so that there exists $P$ for which $A = P^{-1}\operatorname{diag}(\lambda_1, \lambda_2)P$. If $f_P$ denotes the Möbius transform associated to $P$, then this tells that
$$ f_P(p_{n+1}) = f_{\operatorname{diag}(\lambda_1, \lambda_2)}(f_P(p_n)) = \frac{\lambda_1}{\lambda_2}f_P(p_n), $$
so that $q_n = f_P(p_n)$ is simply a geometric sequence.
A: Let $f(x)=\frac{2x}{x+1}$ for $x\in \mathbb R\setminus \{-1\}$. Then, $f'(x)>0$ for any $x\neq-1$ which implies that $f$ is increasing. Hence, the sequence $p_{n+1}=f(p_n)$ is also increasing. Hence, it either converges or goes to infinity. However, $$f(x)=\frac{2x}{x+1}\le \frac{2x}{x}=2$$ if $x<-1$ or $x>0$ and $f(x)<0$ for  $-1<x<0$, so $f(x)$ is bounded.
